Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a^3+b^3+c^3-3abc=\left(a+b\right)^3+c^3-3a^2b-3ab^2-3abc=\left(a+b+c\right)^3\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b\right)-3abc\)
\(=\left(a+b+c\right)\left(a^2+b^2+2ab-ac-bc+c^2\right)-3ab\left(a+b+c\right)=\left(a+b+c\right)\left(a^2+b^2+2ab-ac-bc+c^2-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)
Nhưng theo mình thấy a^3+b^3+c^3 không thể đổi thành (a+b+c)^3
\(a^3+b^3+c^3-3abc=\left(a+b\right)^3+c^3-3a^2b-3ab^2-3abc\)
\(=\left(a+b+c\right)^3\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b\right)-3abc\)
\(=\left(a+b+c\right)\left(a^2+b^2+2ab-ac-bc+c^2\right)-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+2ab-ac-bc+c^2-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)
A = a3 + b3 +c3 -3abc thành nhân tử.
Lời giải:
Từ (a+b)3= a3 + 3a2b +3ab2 + b3
= a3 + b3 + 3ab (a+b)
Ta suy ra: a3 + b3 = (a+b)3 - 3ab (a+b) (1)
áp dụng hằng đẳng thức (1) vào giải bài toán ta có:
A = (a3 + b3) + c3 - 3abc
= (a+b)3 - 3ab (a+b) + c3 - 3abc
= (a+b)3 + c3 - 3ab (a+b) - 3abc
= (a+b+c) (a2 +2ab + b2 -ac - bc + c2 - 3ab)
= (a+b+c) (a2+ b2 +c2 -ab - bc - ac) (*)
a^3 +b^3+c^3-3abc
=a^3+3a^2b+3ab^2+b^3-3a^2b-3ab^2+c^3-3abc
=(a+b)^3+c^3-3ab(a+b+c)
= (a+b+c)((a+b)^2+(a+b)c+c^2)-3ab(a+b+c)
=(a+b+c)(a^2+2ab+b^2+ac+bc+c^-3ab)
=(a+b+c)(a^2+b^2+c^2+ab+bc+ac)
\(a^3+b^3+c^3-3abc\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
\(VT=a^3+b^3+c^3-3abc=\left(a+b\right)^3-3a^2b-3ab^2-3abc+c^2\)
\(=\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2-3ab\right]\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=VP\left(đpcm\right)\)
b) Xét VP ta có :
\(\left(a+b+c\right)\cdot\left(a^2+b^2+c^2-ab-bc-ca\right)\)
\(=a^3+ab^2+ac^2-ab^2-abc-ca^2+ba^2+b^3+bc^2-ab^2-bc^2-abc+ca^2+cb^2+c^3-abc-bc^2-c^2a\)
\(=a^3+b^3+c^3-abc-abc-abc\)
\(=a^3+b^3+c^3-3abc\)
\(=VT\)
Vậy đẳng thức đã được Cm
Bài 1:
a) \(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=\left[\left(a+b+c\right)^3-a^3\right]-\left(b^3+c^3\right)\)
\(=\left(a+b+c-a\right)\left[\left(a+b+c\right)^2+\left(a+b+c\right)a+a^2\right]-\left(b+c\right)\left(b^2-bc+c^2\right)\)
\(=\left(b+c\right)\left(a^2+b^2+c^2+2ab+2bc+2ac+a^2+ab+ac+a^2\right)-\left(b+c\right)\left(b^2-bc+c^2\right)\)
\(=\left(b+c\right)\left(3a^2+3ab+3ac+2bc+b^2+c^2\right)-\left(b+c\right)\left(b^2-bc+c^2\right)\)
\(=\left(b+c\right)\left(3a^2+3ab+3ac+2bc+b^2+c^2-b^2+bc-c^2\right)\)
\(=\left(b+c\right)\left(3a^2+3ab+3ac+3bc\right)\)
\(=3\left(b+c\right)\left(a^2+ab+ac+bc\right)\)
\(=3\left(b+c\right)\left[a\left(a+b\right)+c\left(a+b\right)\right]\)
\(=3\left(b+c\right)\left(a+b\right)\left(a+c\right)\)
b) \(a^3+b^3+c^3-3abc\)
\(=\left(a+b\right)^3+c^3-3a^2b-3ab^2-3abc\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b\right)-3abc\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)
Bài 2:
Từ câu 1b ta đã chứng minh được:
\(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
Thay a + b + c = 0 vào ta được
\(a^3+b^3+c^3-3abc=0\left(a^2+b^2+c^2-ab-bc-ca\right)\)
\(\Rightarrow a^3+b^3+c^3-3abc=0\)
\(\Rightarrow a^3+b^3+c^3=3abc\)
Có: a3 + b3+ c3- 3abc
= (a+b)3- 3a2b - 3ab2- 3abc + c3
=(a+b)3 +c3 - 3ab.(a+b+c)
=(a + b + c). [(a+b)2 - (a+b).c+c2) - 3ab.(a+b+c)
=(a + b + c). ( a2 + 2ab + b2 - ac - bc + c2 - 3ab.(a + b + c)
=(a + b + c). ( a2 + 2ab + b2 - ac - bc + c2 -3ab)
=(a + b + c).( a2 + b2 + c2 - ab - bc - ca)
=>đpcm
chúc bạn học tốt
xét VT = \(a^3+b^3+c^3-3abc\)
nhận xét \(\left(a+b\right)^3=a^3+b^3+3a^2b+3ab^2\)
\(\Rightarrow a^3+b^3=\left(a+b\right)^3-3a^2b-3ab^2\)
thay vào vế trái ta có
\(\left(a+b\right)^3+c^3-3a^2b-3ab^2-3abc\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ac-bc-ab\right)=VP\left(dpcm\right)\)
a+b+c=0
=>(a+b+c)3=0
=>a3+b3+c3+3a2b+3ab2+3b2c+3bc2+3a2c+3ac2+6abc=0
=>a3+b3+c3+(3a2b+3ab2+3abc)+(3b2c+3bc2+3abc)+(3a2c+3ac2+3abc)-3abc=0
=>a3+b3+c3+3ab(a+b+c)+3bc(a+b+c)+3ac(a+b+c)=3abc
Do a+b+c=0
=>a3+b3+c3=3abc(ĐPCM)